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Physics 111: Lecture 4
Today’s Agenda


Recap of centripetal acceleration
Newton’s 3 laws
 How and why do objects move?
 Dynamics
Physics 111: Lecture 4, Pg 1
Review: Centripetal Acceleration

UCM results in acceleration:
Magnitude:
a = v2 / R =  R
Direction:
- r^ (toward center of circle)
v=R
Useful stuff:
R
f = rotations / sec
T=1/f
ω = 2 / T = 2 f = rad/sec
a

Physics 111: Lecture 4, Pg 2
Physics 111: Lecture 4, Pg 3
Dynamics

Isaac Newton (1643 - 1727) published Principia Mathematica
in 1687. In this work, he proposed three “laws” of motion:
Law 1: An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
Law 2: For any object, FNET = F = ma
Law 3: Forces occur in pairs: FA ,B = - FB ,A
(For every action there is an equal and opposite reaction.)
Physics 111: Lecture 4, Pg 4
Newton’s First Law
1. Dishes
2. Monkey

An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
If no forces act, there is no acceleration.

The following statements can be thought of as the definition
of inertial reference frames.
An IRF is a reference frame that is not accelerating (or
rotating) with respect to the “fixed stars”.
If one IRF exists, infinitely many exist since they are
related by any arbitrary constant velocity vector!
Physics 111: Lecture 4, Pg 5
Is Urbana a good IRF?


Ice
puck
Is Urbana accelerating?
YES!
Urbana is on the Earth.
The Earth is rotating.
What is the centripetal acceleration of Urbana?
4 sec,
2
T
=
1
day
=
8.64
x
10
2
v
 2 
aU 
  2R    R
R ~ RE = 6.4 x 106 meters .
T 
R




Plug this in: aU = .034 m/s2 ( ~ 1/300 g)
Close enough to 0 that we will ignore it.
Urbana is a pretty good IRF.
Physics 111: Lecture 4, Pg 6
Newton’s Second Law

For any object, FNET = F = ma.
The acceleration a of an object is proportional to the
net force FNET acting on it.
The constant of proportionality is called “mass”, denoted
m.
» This is the definition of mass.
» The mass of an object is a constant property of that
object, and is independent of external influences.

Force has units of [M]x[L / T2] = kg m/s2 = N (Newton)
Physics 111: Lecture 4, Pg 7
Newton’s Second Law...

What is a force?
A Force is a push or a pull.
A Force has magnitude & direction (vector).
Adding forces is like adding vectors.
a
a
F1
F1
FNET = ma
FNET
F2
F2
Physics 111: Lecture 4, Pg 8
Newton’s Second Law...

Components of F = ma :
FX = maX
FY = maY
FZ = maZ

Suppose we know m and FX , we can solve for aX and
apply the things we learned about kinematics over the last
few weeks:
1
x = x0 + v 0 x t + ax t 2
2
v x = v 0 x + ax t
Physics 111: Lecture 4, Pg 9
Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass m = 100 kg) across
a sheet of ice (horizontal & frictionless). He applies a force
of 50 N in the i direction. If the box starts at rest, what is its
speed v after being pushed a distance d = 10 m?
v=0
F
m
a
i
Physics 111: Lecture 4, Pg 10
Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass m = 100 kg) across
a sheet of ice (horizontal & frictionless). He applies a force
of 50 N in the i direction. If the box starts at rest, what is its
speed v after being pushed a distance d = 10m ?
v
F
m
a
i
d
Physics 111: Lecture 4, Pg 11
Example: Pushing a Box on Ice...

Start with F = ma.
a = F / m.
Recall that v2 - v02 = 2a(x - x0 )
So
v2
v
= 2Fd / m
(Lecture 1)
2 Fd
m
v
F
m
a
i
d
Physics 111: Lecture 4, Pg 12
Example: Pushing a Box on Ice...
v

2 Fd
m
Plug in F = 50 N, d = 10 m, m = 100 kg:
Find v = 3.2 m/s.
v
F
m
a
i
d
Physics 111: Lecture 4, Pg 13
Lecture 4, Act 1
Force and acceleration

A force F acting on a mass m1 results in an acceleration a1.
The same force acting on a different mass m2 results in an
acceleration a2 = 2a1.
m1
F

a1
F
m2
a2 = 2a1
If m1 and m2 are glued together and the same force F acts
on this combination, what is the resulting acceleration?
F
(a)
2/3 a1
m1
m2
(b) 3/2 a1
a=?
(c)
3/4 a1
Physics 111: Lecture 4, Pg 14
Lecture 4, Act 1
Force and acceleration
m1
F
m2
a = F / (m1+ m2)

Since a2 = (1/2) a1 for the same applied force, m2 = (1/2)m1 !
m1+ m2 = 3m2 /2

So a = (2/3)F / m1
but F/m = a
a = 2/3 a1
(a)
2/3 a1
(b) 3/2 a1
(c)
3/4 a1
Physics 111: Lecture 4, Pg 15
Forces

We will consider two kinds of forces:
Contact force:
» This is the most familiar kind.
I push on the desk.
 The ground pushes on the chair...

Action at a distance:
» Gravity
» Electricity
Physics 111: Lecture 4, Pg 16
Contact forces:

Objects in contact exert forces.

Convention: Fa,b means
“the force acting on a due to b”.

So Fhead,thumb means “the force on
the head due to the thumb”.
Fhead,thumb
Physics 111: Lecture 4, Pg 17
Action at a distance

Gravity:
Physics 111: Lecture 4, Pg 18
Gravitation
(Courtesy of Newton)


Newton found that amoon / g = 0.000278
and noticed that RE2 / R2 = 0.000273
amoon
g
R

RE
This inspired him to propose the
Universal Law of Gravitation:
|FMm |= GMm / R2
where G = 6.67 x 10 -11 m3 kg-1 s-2
Physics 111: Lecture 4, Pg 19
Gravity...

The magnitude of the gravitational force F12 exerted on an
object having mass m1 by another object having mass m2
a distance R12 away is:
F12  G

m1 m2
2
R12
The direction of F12 is attractive, and lies along the line
connecting the centers of the masses.
m1
F12
F21
m2
R12
Physics 111: Lecture 4, Pg 20
Gravity...

Near the Earth’s surface:
R12 = RE
» Won’t change much if we stay near the Earth's surface.
» i.e. since RE >> h, RE + h ~ RE.
h
m
Fg
Fg  G
ME m
RE2
M
RE
Physics 111: Lecture 4, Pg 21
Gravity...

Near the Earth’s surface...

So |Fg| = mg = ma

a=g
Where: g  G
Leaky Cup
 ME 
ME m
Fg  G 2  m G 2 
RE
 RE 

=g
All objects accelerate with
acceleration g, regardless of
their mass!
ME
2

9
.
81
m
/
s
RE2
Physics 111: Lecture 4, Pg 22
Example gravity problem:

What is the force of gravity exerted by the earth on a typical
physics student?
Typical student mass m = 55kg
g = 9.8 m/s2.
Fg = mg = (55 kg)x(9.8 m/s2 )
Fg = 539 N

The force that gravity exerts on any object is
called its Weight
W = 539 N
Fg
Physics 111: Lecture 4, Pg 23
Lecture 4, Act 2
Force and acceleration


Suppose you are standing on a bathroom scale in 141 Loomis
and it says that your weight is W. What will the same scale say
your weight is on the surface of the mysterious Planet X ?
You are told that RX ~ 20 REarth and MX ~ 300 MEarth.
(a)
0.75 W
(b)
1.5 W
(c)
2.25 W
E
X
Physics 111: Lecture 4, Pg 24
Lecture 4, Act 2
Solution


The gravitational force on a person
of mass m by another object (for instance
a planet) having mass M is given by:
Ratio of weights = ratio of forces:
MX m
R X2

M m
G E2
RE
F G
Mm
R2
W X FX

WE FE
G
M
 X
ME
R 
 E 
 RX 
2
2
WX
1
 300     .75
 20 
WE
Physics 111: Lecture 4, Pg 25
Newton’s Third Law:

Newton’s
Sailboard
Forces occur in pairs: FA ,B = - FB ,A.
For every “action” there is an equal and opposite “reaction”.

We have already seen this in the case of gravity:
m1
m2
F12
F21
F12  G
m1 m2
 F21
2
R12
R12
Physics 111: Lecture 4, Pg 26
Newton's Third Law...

2 Skateboards
FA ,B = - FB ,A. is true for contact forces as well:
Fm,w
Fw,m
Ff,m
Fm,f
Physics 111: Lecture 4, Pg 27
Example of Bad Thinking

Since Fm,b = -Fb,m, why isn’t Fnet = 0 and a = 0 ?
Fm,b
Fb,m
a ??
ice
Physics 111: Lecture 4, Pg 28
Example of Good Thinking

Consider only the box as the system!
Fon box = mabox = Fb,m
Free Body Diagram (next time).
Fm,b
Fb,m
abox
ice
Physics 111: Lecture 4, Pg 29
Lecture 4, Act 3
Newton’s 3rd Law

Two blocks are stacked on the ground. How many action-reaction
pairs of forces are present in this system?
(a) 2
a
(b) 3
b
(c) 4
Physics 111: Lecture 4, Pg 30
Lecture 4, Act 3
Solution:
a
Fa,E
b
a
b
Fb,E
a
a
b
b
Fb,a
Fa,b
Fb,g
Fg,b
FE,a
FE,b
(c) 4
Physics 111: Lecture 4, Pg 31
Recap of today’s lecture
Extinguisher
Cart

Newton’s 3 Laws:
(Text: 4-1 to 4-5)
Law 1: An object subject to no external forces is at rest or
moves with a constant velocity if viewed from an inertial
reference frame.
(Text: 4-1)
Law 2: For any object, FNET = F = ma
(Text: 4-2 & 4-3)
Law 3: Forces occur in pairs: FA ,B = - FB ,A. (Text: 4-4 & 4-5)

Look at Textbook problems Chapter 4: # 3, 5, 7, 19
Physics 111: Lecture 4, Pg 32